An object of mass 0.5 kg is executing simple harmonic motion. It amplitude is 5 cm and time period (T) is 0.2 s. What will be the potential energy of the object at an instant t = T/4s starting from mean position. Assume that the initial phase of the oscillation is zero.

Solution

Physics sir · 4 steps

A quarter period after leaving the mean, it sits at the extreme, so all its energy is potential.

1. Where is it. Starting from the mean position with zero phase, x = A sin ω t. At t = T/4, ω t = 2π/T × T/4 = π/2, so x = A. The object is at the extreme position.
2. Angular frequency. ω = 2π/T = 2π/0.2 = 10π rad s⁻¹. Convert the amplitude to metres: A = 5 cm = 0.05 m.

   U = 1/2mω² x² = 1/2mω² A²

3. Put in numbers. U = 1/2 × 0.5 × (10π)² × (0.05)² = 0.25 × 100π² × 0.0025 = 0.0625π². With π² ≈ 9.87, U ≈ 0.617 J.
4. So. The potential energy at t = T/4 is about 0.62 J, which is option (a).

The idea to keep. In SHM, find the position first. At T/4 from the mean it is at the extreme, where U = 1/2mω²A² equals the total energy. Always put the amplitude in metres.

Answer from the NTA final answer key. Solution by PrepWiser.